Question

You drop a single coffee filter of mass 1.4 grams from a very tall building, and it takes 49 seconds to reach the ground. In a small fraction of that time the coffee filter reached terminal speed. (a) What was the upward force of the air resistance while the coffee filter was falling at terminal speed? Fair = N (b) Next you drop a stack of 5 of these coffee filters. What was the upward force of the air resistance while this stack of coffee filter was falling at terminal speed? Fair = N (c) Again assuming again that the stack reaches terminal speed very quickly, about how long will the stack of coffee filters take to hit the ground? (Hint: Consider the relation between speed and the force of air resistance.) Fall time is approximately s Additional Materials Section 7.10

Answer 1

a) 0.014N

b) 0.069N

c) 22 seconds

Step-by-step explanation:

We are given:

t = 49 seconds

`F = 1.4g = 1.4*10^-^3 Kg`

let's take
`g = 9.8m/s^2`

a) At terminal speed, net force will be zero.

Therefore,

`F_a_i_r = m * g`

`F_a_i_r = (1.4*10^-^3kg)(9.8 m/s^2)= 0.014N`

2. Since 5 coffee filters were dropped, we assume the gravity force and mass will increase by 5, so we have:

`F_a_i_r = 5 * 1.4*10^-^3 *9.8`

= 0.069N

3. At terminal velocity, drag force will be a function of V²

Therefore,

`mg=(1)/(2)kv^2`

Terminal velocity will now be
`= √(m)`

We are told it takes 49 seconds for 1 filter to hit the ground, thus for 5 filters to hit the grround we have:

`F_d_r_a_g = (49)/(√(5))`

= 21.9 => 22 seconds

It will take approximately 22 seconds for a stack of 5 filters to hit the ground